(15 points ) The City of Melbourne currently operates a reverse osmosis desalina

Question

(15 points ) The City of Melbourne currently operates a reverse osmosis desalination plant to supplement the existing water supply. Reverse osmosis separates a feed water stream into a product stream and a high-salinity waste stream. The groun 1800 streem no more than 3000 mg/L. If the plant capacity is 8.0 MGD of 2 dwater of waste Will be produced? (Hint: write a material balance on salt. feed is expected to contain of salt. The product water must contain no more than 200 mg/t, and the waste MGD

Explanation / Answer

Hi,

Hope it helps.

PLEASE GIVE THUMBS UP.

Thanks.

It is said that the two components are product water and waste water.

Of the 1800 mg/l of salt we have 1600 mg/l of waste salt as product water can contain a maximum of 200 mg/l of salt.

Plant capacity = 8.0 MGD of product , i.e 8.0 MGD contains 200 mg/l of salt.

Therefore for 1600 mg/l of salt ,

the MGD waster produced = 8.0 * 1600 /200

= 8.0*8.0

= 64.0 MGD

In terms of material balance,

it will be,

[8.0 * 200 ] + [X * 1600] = [X + 8.0 ]* [1800]

which gives X = 64.0 MGD.

Now please GIVE THUMBS UP Sir.

Thanks.

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